ZOJ 3661 Palindromic Substring(回文树)

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In the kingdom of string, people like palindromic strings very much. They like only palindromic strings and dislike all other strings. There is a unified formula to calculate the score of a palindromic string. The score is calculated by applying the following three steps.

1. Since a palindromic string is symmetric, the second half(excluding the middle of the string if the length is odd) is got rid of, and only the rest is considered. For example, "abba" becomes "ab", "aba" becomes "ab" and "abacaba" becomes "abac".
2. Define some integer values for ‘a‘ to ‘z‘.
3. Treat the rest part as a 26-based number M and the score is M modulo 777,777,777.

However different person may have different values for ‘a‘ to ‘z‘. For example, if ‘a‘ is defined as 3, ‘b‘ is defined as 1 and c is defined as 4, then the string "accbcca" has the score (3×26³+4×26²+4×26+1) modulo 777777777=55537.

One day, a very long string S is discovered and everyone in the kingdom wants to know that among all the palindromic substrings of S, what the one with the K-th smallest score is.

Input

The first line contains an integer T(1 ≤ T ≤ 20), the number of test cases.

The first line in each case contains two integers n, m(1 ≤ n ≤ 100000, 1 ≤ m ≤ 20) where n is the length of S and m is the number of people in the kingdom. The second line is the string S consisting of only lowercase letters. The next m lines each containing 27 integers describes a person in the following format.

K_i va vb ... vz

where va is the value of ‘a‘ for the person, vb is the value of ‘b‘ and so on. It is ensured that the K_i-th smallest palindromic substring exists and va, vb, ..., vz are in the range of [0, 26). But the values may coincide.

Output

For each person, output the score of the K-th smallest palindromic substring in one line. Print a blank line after each case.

Sample Input

3
6 2
abcdca
3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
7 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
4 10
zzzz
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
51 4
abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba
1 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
25 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
26 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
76 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1

Sample Output

1
620

14
14
14
14
14
14
14
378
378
378

0
9
14
733665286

回文树
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;
typedef long long int LL;
const int maxn=1e5+5;
const int mod=777777777;
char str[maxn];
int n,m;
LL k;
int a[26];
LL pow(int x)
{
	LL sum=1;
	LL n=26;
	for(x;x;x>>=1)
	{
		if(x&1)
			sum=(sum*n)%mod;
		n=(n*n)%mod;
	}
	return sum;
}
struct Node
{
	LL num;
	LL sum;
}c[maxn];
int cmp(Node a,Node b)
{
	return a.sum<b.sum;
}
struct Tree
{
	int next[maxn][26];
	int fail[maxn];
	LL num[maxn];
	int cnt[maxn];
	int len[maxn];
	int s[maxn];
	int last,p,n;
	int new_node(int x)
	{
		memset(next[p],0,sizeof(next[p]));
		cnt[p]=0;
		num[p]=0;
		len[p]=x;
		return p++;
	}
	void init()
	{
		p=0;
		new_node(0);
		new_node(-1);
		last=0;
		n=0;
		s[0]=-1;
		fail[0]=1;
	}
	int get_fail(int x)
	{
		while(s[n-len[x]-1]!=s[n])
			x=fail[x];
		return x;
	}
	int add(int x)
	{
		x-='a';
		s[++n]=x;
		int cur=get_fail(last);
		if(!(last=next[cur][x]))
		{
			int now=new_node(len[cur]+2);
			fail[now]=next[get_fail(fail[cur])][x];
			next[cur][x]=now;
			num[now]=(num[cur]+((LL)pow((len[cur]+1)/2)*a[x])%mod)%mod;
			last=now;
		}
		cnt[last]++;
		return 1;
	}
	void count()
	{
		for(int i=p-1;i>=0;i--)
			cnt[fail[i]]+=cnt[i];
	}
	void fun()
	{
		count();
		int cot=0;
		for(int i=2;i<p;i++)
		{
			c[cot].num=cnt[i];
			c[cot++].sum=num[i];
		}
		sort(c,c+cot,cmp);
		int i;
		for( i=0;i<cot;i++)
		{
			if(k>c[i].num)
			{
				k-=c[i].num;
			}
			else
				break;
		}
        printf("%d\n",c[i].sum);
	}
}tree;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
        scanf("%d%d",&n,&m);
		scanf("%s",str);
		for(int i=1;i<=m;i++)
		{
			scanf("%lld",&k);
			for(int j=0;j<26;j++)
				scanf("%d",&a[j]);
			tree.init();
			for(int j=0;j<n;j++)
			{
                 tree.add(str[j]);
			}
			tree.fun();
        }
		cout<<endl;
	}
	return 0;
}

ZOJ 3661 Palindromic Substring(回文树)

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原文地址：http://blog.csdn.net/dacc123/article/details/51367559